Question

Question #fef55

Answer 1

The reaction's percent yield is `"90.5%"`.

So, start with the balanced chemical equation for the decomposition of `CaCO_3`

`CaCO_(3(s)) -> CaO_((s)) + CO_(2(g))`

Notice that you've got `"1:1"` mole ratios between all the compounds involved in the reaction. This is important because you'll use the mole ratio `CaCO_3` has with `CaO` to determine what the theoretical yield of the reaction is.

The number of `CaCO_3` moles you have is

`2.00 * 10^(3)"g" * ("1 mole")/("100.1 g") = "20.0 moles"`

This means that the mass of `CaO` produced if the reaction had a 100% yield would be

`"20.0 moles" * ("58.1 g")/("1 mole") = "1162 g" = 1.16 * 10^(3)"g"`

Since percent yield is defined as actual yield divided by theoretical yield and multiplied by 100, you'd get

`"% yield" = "actual"/"theoretic" * 100 = (1.05 * 10^(3)"g")/(1.16 * 10^(3)"g") * 100 = "90.5%"`

You can double check the result by using moles instead of grams. You know that you have `"20 moles"` of `CaO` produced in theory. The actual number of moles of `CaO` produced is

`1.05 * 10^(3)"g" * ("1 mole")/("58.1 g") = "18.1 moles"`

Therefore, the percent yield is one again

`"% yield" = ("18.1 moles CaO")/("20.0 moles CaO") * 100 = "90.5%"`